Question

The lifespan of a fruit fly is normally distributed with an average of 40 days and a standard deviation of 7.4 days. a) What is the probability that a randomly selected fruit fly lives longer than 50 days? b) If a fruit fly lives longer than 75% of all other fruit flies, how long did it live? 1 c) If we take a sample of 80 fruit flies, what is the probability the sample mean is within 1 day of the population mean?

Answer 1

Let X be the lifespan of a fruit fly

X~ Normal ( 40, 7.4)

a) P( X > 50) = P( > )

= P( z > 1.35)

= 1- P( z < 1.35)

= 1- 0.91149

= 0.08851

b) Here, we need to find the 75th percentile

P( Z < z ) =0.75

P ( Z < 0.674 ) =0.75

z = 0.674

= 0.674

   = 0.674

X = 40 + ( 0.674 * 7.4)

X = 44.9876

c) Sample of 80 fruit flies are taken , n= 80

Let be the mean lifespan of the 80 fruit flies

   ~ Normal ( 40, )

P( 39 < < 41 ) = P ( < < )

= P( -1.21 < z < 1.21)

= P( z < 1.21) - P( z < -1.21)

= P( z < 1.21) - 1 + P ( z < 1.21)

= 2 * P(z < 1.21) -1

= 2 * 0.88686 -1

= 0.77372